Let f(x) be a polynomial of degree 3 such that f(-1) = 10, f(1) = -6, f(x) has a critical point at x = -1 and f'(x) has a critical point at x = 1. Then f(x) has a local minima at x = ____

Answer

To solve the problem, we need to find the local minima of the polynomial \( f(x) \) given the conditions provided. Let's go through the solution step by step.

### Step 1: Define the Polynomial
Since \( f(x) \) is a polynomial of degree 3, we can express it in the general form:
\[ f(x) = ax^3 + bx^2 + cx + d \]

### Step 2: Use the Given Conditions
We have the following conditions:
1. \( f(-1) = 10 \)
2. \( f(1) = -6 \)
3. \( f'(-1) = 0 \) (critical point at \( x = -1 \))
4. \( f''(1) = 0 \) (critical point of the derivative at \( x = 1 \))

### Step 3: Calculate the Derivative
First, we calculate the first derivative of \( f(x) \):
\[ f'(x) = 3ax^2 + 2bx + c \]
Next, we calculate the second derivative:
\[ f''(x) = 6ax + 2b \]

### Step 4: Apply the Conditions
Using the conditions, we can set up equations.

1. From \( f(-1) = 10 \):
\[ f(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = 10 \quad \text{(Equation 1)} \]

2. From \( f(1) = -6 \):
\[ f(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = -6 \quad \text{(Equation 2)} \]

3. From \( f'(-1) = 0 \):
\[ f'(-1) = 3a(-1)^2 + 2b(-1) + c = 3a - 2b + c = 0 \quad \text{(Equation 3)} \]

4. From \( f''(1) = 0 \):
\[ f''(1) = 6a(1) + 2b = 6a + 2b = 0 \quad \text{(Equation 4)} \]

### Step 5: Solve the Equations
From Equation 4:
\[ 2b = -6a \implies b = -3a \]

Substituting \( b = -3a \) into Equations 1, 2, and 3:

**Substituting into Equation 1:**
\[ -a + (-3a) - c + d = 10 \implies -4a - c + d = 10 \quad \text{(Equation 5)} \]

**Substituting into Equation 2:**
\[ a + (-3a) + c + d = -6 \implies -2a + c + d = -6 \quad \text{(Equation 6)} \]

**Substituting into Equation 3:**
\[ 3a - 2(-3a) + c = 0 \implies 3a + 6a + c = 0 \implies 9a + c = 0 \implies c = -9a \quad \text{(Equation 7)} \]

### Step 6: Substitute \( c \) into Equations 5 and 6
**Substituting \( c = -9a \) into Equation 5:**
\[ -4a - (-9a) + d = 10 \implies 5a + d = 10 \quad \text{(Equation 8)} \]

**Substituting \( c = -9a \) into Equation 6:**
\[ -2a - 9a + d = -6 \implies -11a + d = -6 \quad \text{(Equation 9)} \]

### Step 7: Solve for \( a \) and \( d \)
Now we have two equations (8 and 9):
1. \( 5a + d = 10 \)
2. \( -11a + d = -6 \)

Subtracting these equations:
\[ (5a + d) - (-11a + d) = 10 - (-6) \implies 16a = 16 \implies a = 1 \]

Substituting \( a = 1 \) back into Equation 8:
\[ 5(1) + d = 10 \implies d = 5 \]

### Step 8: Find \( b \) and \( c \)
Using \( a = 1 \):
\[ b = -3(1) = -3, \quad c = -9(1) = -9 \]

### Step 9: Write the Polynomial
Thus, the polynomial is:
\[ f(x) = x^3 - 3x^2 - 9x + 5 \]

### Step 10: Find Critical Points
To find local minima, we need to find the critical points of \( f'(x) \):
\[ f'(x) = 3x^2 - 6x - 9 \]
Setting \( f'(x) = 0 \):
\[ 3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \]
Factoring:
\[ (x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1 \]

### Step 11: Determine Local Minima
To determine the nature of these critical points, we can use the second derivative test:
\[ f''(x) = 6x - 6 \]
Evaluating at \( x = 3 \):
\[ f''(3) = 6(3) - 6 = 12 > 0 \quad \text{(local minima)} \]
Evaluating at \( x = -1 \):
\[ f''(-1) = 6(-1) - 6 = -12 < 0 \quad \text{(local maxima)} \]

### Conclusion
Thus, \( f(x) \) has a local minima at \( x = 3 \).